If `|alpha|lt1,|beta|lt1` `1-alpha+alpha^(2)-alpha^(3)+ . . . . ."to "oo=s_(1)` `1-beta+beta^(2)-beta^(3)+ . . . ."to "oo=s_(2)`, then `1-alphabeta+alpha^(2)beta^(2)+ . . . . ."to "oo` equals

Options

A. `s_(1)s_(2)`
B. `(s_(1)s_(2))/(1+s_(1)s_(2))`
C. `(s_(1)s_(2))/(1-s_(1)-s_(2)+2s_(1)s_(2))`
D. `(1)/(1+s_(1)s_(2))`

Answer

C

Explanation

We have,
`1-alpha+alpha^(2)-alpha^(3)+ . . . . ."to "oo=s_(1)`
`and,1-beta+beta^(2)-beta^(3)+ . . . ."to "oo=s_(2)`
`rArr" "(1)/(a+alpha)=s_(1)and(1)/(1+beta)=s_(2)`
`rArr" "alpha=(1)/(s_(1))-1andbeta=(1)/(s_(2))-1`
`:." "1-alphabeta+alpha^(2)beta^(2)-a^(3)beta^(3)+ . . ."to "oo`
`=(1)/(1+alphabeta)+(1)/(1+((1)/s_(1)-1)((1)/(s_(2))-1))=(s_(1)s_(2))/(1-(s_(1)+s_(2))+2s_(1)s_(2))`